A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of 60 with vertical, then maximum induced EMF between the point of suspension and point of oscillation is mV. (Take g = 10 m/s2)
Answer
100
Solution
The EMF is maximum when the angular speed is maximum, i.e. at the lowest point.
Energy conservation (l = 0.1 m): v2 = 2gl(1 - 60 ) = 2(10)(0.1) ( 12 ) = 1
v = 1 m/s, = vl = 10.1 = 10 rad/s
EMF across a rod of length l rotating about one end in a perpendicular field B:
= 12B l2 = 12(2)(10)(0.1)2 = 0.1 V
max = 100 mV
Answer: 100